Termination w.r.t. Q proof of /tmp/tmpeGeudb/Ex6_9_Luc02c

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 PisEmptyProof (⇔)

    ↳18 TRUE

  ↳19 QDP

    ↳20 QDPOrderProof (⇔)

    ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 QDPOrderProof (⇔)

    ↳27 QDP

    ↳28 PisEmptyProof (⇔)

    ↳29 TRUE

  ↳30 QDP

    ↳31 QDPOrderProof (⇔)

    ↳32 QDP

    ↳33 QDPOrderProof (⇔)

    ↳34 QDP

    ↳35 QDPOrderProof (⇔)

    ↳36 QDP

    ↳37 QDPOrderProof (⇔)

    ↳38 QDP

    ↳39 PisEmptyProof (⇔)

    ↳40 TRUE

  ↳41 QDP

    ↳42 QDPOrderProof (⇔)

    ↳43 QDP

    ↳44 QDPOrderProof (⇔)

    ↳45 QDP

    ↳46 PisEmptyProof (⇔)

    ↳47 TRUE

  ↳48 QDP

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVE(2nd(cons1(X, cons(Y, Z)))) → MARK(Y)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
ACTIVE(2nd(cons(X, X1))) → 2ND(cons1(X, X1))
ACTIVE(2nd(cons(X, X1))) → CONS1(X, X1)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))
MARK(2nd(X)) → 2ND(mark(X))
MARK(2nd(X)) → MARK(X)
MARK(cons1(X1, X2)) → ACTIVE(cons1(mark(X1), mark(X2)))
MARK(cons1(X1, X2)) → CONS1(mark(X1), mark(X2))
MARK(cons1(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(from(X)) → FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
2ND(mark(X)) → 2ND(X)
2ND(active(X)) → 2ND(X)
CONS1(mark(X1), X2) → CONS1(X1, X2)
CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(active(X1), X2) → CONS1(X1, X2)
CONS1(X1, active(X2)) → CONS1(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
FROM(mark(X)) → FROM(X)
FROM(active(X)) → FROM(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 10 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(active(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1) = S(x1)
active(x1) = active(x1)
mark(x1) = x1

Recursive path order with status [RPO].
Precedence:

trivial

Status:

active₁: multiset
S₁: multiset

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(mark(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Precedence:

mark₁ > S₁

Status:

mark₁: multiset
S₁: multiset

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FROM(active(X)) → FROM(X)
FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FROM(active(X)) → FROM(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FROM(x1) = FROM(x1)
active(x1) = active(x1)
mark(x1) = x1

Recursive path order with status [RPO].
Precedence:

trivial

Status:

active₁: multiset
FROM₁: multiset

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FROM(mark(X)) → FROM(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Precedence:

mark₁ > FROM₁

Status:

mark₁: multiset
FROM₁: multiset

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(17) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(18) TRUE

(19) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(active(X1), X2) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = CONS(x1)
mark(x1) = x1
active(x1) = active(x1)

Recursive path order with status [RPO].
Precedence:

trivial

Status:

active₁: multiset
CONS₁: multiset

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(mark(X1), X2) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = CONS(x1)
mark(x1) = mark(x1)
active(x1) = active

Recursive path order with status [RPO].
Precedence:

trivial

Status:

active: multiset
CONS₁: [1]
mark₁: multiset

The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(X1, mark(X2)) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = CONS(x1, x2)
mark(x1) = mark(x1)
active(x1) = x1

Recursive path order with status [RPO].
Precedence:

mark₁ > CONS₂

Status:

CONS₂: [1,2]
mark₁: multiset

The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(X1, active(X2)) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Precedence:

active₁ > CONS₂

Status:

active₁: multiset
CONS₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(27) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(28) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(29) TRUE

(30) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(mark(X1), X2) → CONS1(X1, X2)
CONS1(active(X1), X2) → CONS1(X1, X2)
CONS1(X1, active(X2)) → CONS1(X1, X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(31) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS1(active(X1), X2) → CONS1(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS1(x1, x2) = CONS1(x1)
mark(x1) = x1
active(x1) = active(x1)

Recursive path order with status [RPO].
Precedence:

trivial

Status:

active₁: multiset
CONS1₁: multiset

The following usable rules [FROCOS05] were oriented: none

(32) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(mark(X1), X2) → CONS1(X1, X2)
CONS1(X1, active(X2)) → CONS1(X1, X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS1(mark(X1), X2) → CONS1(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS1(x1, x2) = CONS1(x1)
mark(x1) = mark(x1)
active(x1) = active

Recursive path order with status [RPO].
Precedence:

trivial

Status:

active: multiset
CONS1₁: [1]
mark₁: multiset

The following usable rules [FROCOS05] were oriented: none

(34) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(X1, active(X2)) → CONS1(X1, X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS1(X1, mark(X2)) → CONS1(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS1(x1, x2) = CONS1(x1, x2)
mark(x1) = mark(x1)
active(x1) = x1

Recursive path order with status [RPO].
Precedence:

mark₁ > CONS1₂

Status:

CONS1₂: [1,2]
mark₁: multiset

The following usable rules [FROCOS05] were oriented: none

(36) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS1(X1, active(X2)) → CONS1(X1, X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS1(X1, active(X2)) → CONS1(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Precedence:

active₁ > CONS1₂

Status:

active₁: multiset
CONS1₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(38) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(39) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(40) TRUE

(41) Obligation:

Q DP problem:
The TRS P consists of the following rules:

2ND(active(X)) → 2ND(X)
2ND(mark(X)) → 2ND(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(42) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

2ND(active(X)) → 2ND(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
2ND(x1) = 2ND(x1)
active(x1) = active(x1)
mark(x1) = x1

Recursive path order with status [RPO].
Precedence:

trivial

Status:

active₁: multiset
2ND₁: multiset

The following usable rules [FROCOS05] were oriented: none

(43) Obligation:

Q DP problem:
The TRS P consists of the following rules:

2ND(mark(X)) → 2ND(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(44) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

2ND(mark(X)) → 2ND(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Precedence:

mark₁ > 2ND₁

Status:

2ND₁: multiset
mark₁: multiset

The following usable rules [FROCOS05] were oriented: none

(45) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(46) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(47) TRUE

(48) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MARK(2nd(X)) → ACTIVE(2nd(mark(X)))
ACTIVE(2nd(cons1(X, cons(Y, Z)))) → MARK(Y)
MARK(2nd(X)) → MARK(X)
MARK(cons1(X1, X2)) → ACTIVE(cons1(mark(X1), mark(X2)))
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
MARK(cons1(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(from(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.